116 mg of a compound on vaporisation in a Vi | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given mass of the compound $= 116 \ mg = 116 	imes 10^{-3} \ g$.Volume of air displaced at $S.T.P. = 44.8 \ mL = 44.8 	imes 10^{-3} \ L$.At $S.T

Vedclass · Accepted Answer

$58$

Vedclass · Answer

(C) Given mass of the compound $= 116 \ mg = 116 	imes 10^{-3} \ g$.Volume of air displaced at $S.T.P. = 44.8 \ mL = 44.8 	imes 10^{-3} \ L$.At $S.T.P.$,$1 \ mole$ of any gas occupies $22.4 \ L$ or $22400 \ mL$.Number of moles of the compound $= \frac{	ext{Volume at S.T.P. (mL)}}{22400 \ mL/mol} = \frac{44.8}{22400} = 0.002 \ mol$.Molecular weight $= \frac{	ext{Mass}}{	ext{Moles}} = \frac{116 	imes 10^{-3} \ g}{0.002 \ mol} = 58 \ g/mol$.

$116 \ mg$ of a compound on vaporisation in a Victor Meyer's apparatus displaces $44.8 \ mL$ of air measured at $S.T.P.$ The molecular weight of the compound is

Similar Questions

In the chemical scale,the relative atomic mass of the naturally occurring isotopic mixture of oxygen atoms $({}^{16}O, {}^{17}O, {}^{18}O)$ is taken as:

The specific heat of a metal $M$ is $0.25 \ J \ g^{-1} \ K^{-1}$. Its equivalent weight is $12$. What is its atomic weight?

Atoms or groups of atoms which are electrically charged are known as:

$0.126 \ g$ of an acid is needed to completely neutralise $20 \ mL$ of $0.1 \ N$ $NaOH$ solution. The equivalent weight of the acid is:

Atomic weight of an element is $x$. The actual mass of one atom of that element is

Vedclass Test Series

Exam Paper Generator

Online Exam Module